# Question 9:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin 2x - \tan x = 2\sin x \cos x - \tan x$ | M1 | Uses correct double angle identity for $\sin 2x$; accept $\sin(x+x) = \sin x\cos x + \cos x\sin x$ |
| $= \frac{2\sin x\cos^2 x}{\cos x} - \frac{\sin x}{\cos x}$ | M1 | Uses $\tan x = \frac{\sin x}{\cos x}$ with $\sin 2x = 2\sin x\cos x$; attempts common denominator |
| $= \frac{\sin x}{\cos x}(2\cos^2 x - 1)$ | dM1 | Both M marks scored; uses correct double angle identity for $\cos 2x$ |
| $= \tan x\cos 2x$ | A1* | Fully correct, no errors or omissions; all notation correct |

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan x(\cos 2x - 3\sin x) = 0$ | M1 | $\tan x$ cancelled/factored to give $\cos 2x - 3\sin x = 0$; condone slips |
| $1 - 2\sin^2 x - 3\sin x = 0$ | M1 | Uses $\cos 2x = 1-2\sin^2 x$ to form 3TQ in $\sin x$ |
| $\sin x = \frac{-3\pm\sqrt{17}}{4}$ | M1 | Uses formula/completion of square/GC with inverse sin |
| Two of $x = 16.3°, 163.7°, 0°, 180°$ | A1 | awrt $16.3°$, awrt $163.7°$; or in radians awrt 0.28, 2.86, 0 and $\pi$ |
| All four of $x = 16.3°, 163.7°, 0°, 180°$ | A1 | All four in degrees, no extras inside range $0 \leq x < 360°$; condone $0=0.0$ and $180°=180.0°$ |

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